Problem: Ten points lie on a circle. How many inscribed quadrilaterals can be drawn having these points as vertices?

My Thoughts: I'm trying to make a connection between this problem and one I already know:"Seven points lie on a circle, as shown in the figure. How many inscribed triangles can be drawn having any 3 of these points as vertices?" Let n = the number of points on the circle (n > 2), then we can use the formula [(n - 2)(n - 1)] / 2 to find the number of inscribed triangles that can be drawn using n points as vertices. So with 7 points on the circle, 15 triangles can be formed.

I know that circular permutations/ combinations will not follow the same formulas as regular permutations/ combinations. I tried to sketch out my options, but there got to be so many that it was very confusing. I also started to make a chart as far as listing n = the number of points on the circle, and finding the number of quadrilaterals that could be formed. I have: When n = 4, 1 quad. is formed.When n = 5, 5 quads. are formed.When n = 6, 12 quads are formed.And then when I get to n = 7, my diagram gets too complicated to see.

I am not sure how valid a conjecture on a formula would be with only the first 3 n values to work with. Is anyone aware of a formula that could be used to represent this problem? Or a website that could lead me through an explanation?

gingersnap0819 wrote:Problem: Ten points lie on a circle. How many inscribed quadrilaterals can be drawn having these points as vertices?

My Thoughts: I'm trying to make a connection between this problem and one I already know:

"Seven points lie on a circle, as shown in the figure. How many inscribed triangles can be drawn having any 3 of these points as vertices?" Let n = the number of points on the circle (n > 2), then we can use the formula [(n - 2)(n - 1)] / 2 to find the number of inscribed triangles that can be drawn using n points as vertices. So with 7 points on the circle, 15 triangles can be formed.

What was the reasoning behind the solution for triangles? What thinking generated the "(n - 2)(n - 1)/2" formula? Follow the same logic for quadrilaterals.